Where are the x-intercepts for f(x) = −4cos(x − pi over 2 ) from x = 0 to x = 2π?
@DanJS
Have you considered substituting f(x) = 0?
Yes, I just don't know how to simplify with the "cos" value in the equation; it's really confusing.
You don't simplify cos(x). Why are you trying to do that? You know things about it. −4cos(x − pi/2 ) = 0 -4 is of no consequence for this problem. This leaves: cos(x − pi/2 ) = 0 Where is cos(x) = 0? You answer this question. Tell me EVERYWHERE this is the case.
Hmm, I still don't understand. :/
Simple question. Ignore that there is a different question. Just answer this question. Fundamental knowledge of the cosine function. Where is cos(x) = 0? It is a periodic function. It takes on the value zero infinitely many times. Where are all these times? They are very regular and can be stated rather succinctly.
I'm sorry, I really don't know. @tkhunny
That's not acceptable. I am sure you know that \(\cos(\pi/2) = 0\). I am sure you know that \(\cos(3\pi/2) = 0\). I am relatively sure you know that \(\cos(-\pi/2) = 0\). Those are just three of the infinitely many values where this is so. What are the rest of the values. You MUST know them all in order to have a sufficient understanding of the cosine.
How in the world will I determine any of these functions?
@tkhunny
**I mean values
You must study the properties. Here's a 10 minute video that is helpful. https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/unit-circle-definition-of-trig-functions/v/unit-circle-definition-of-trig-functions-1
the symmetry of things on that circle is useful , like Cos(-angle) = Cos(+angle), x value stays the same... with that you can write cos(theta - pi/2) = cos( pi/2 - theta) the cosine of pi/2 - theta, is the same as the sine value of the angle theta, cosine of an angle is the same as the sine of the angle compliment , like sin(12) = cos(78), if you can remember those kinds of things, then the problem is reduced to figuring when the sin(theta) = 0
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